# How do you differentiate y=sin^-1(1/x)?

Oct 27, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{x \sqrt{{x}^{2} - 1}}$

#### Explanation:

The easiest way is to rewrite $y = {\sin}^{-} 1 \left(\frac{1}{x}\right)$ as $\sin y = \frac{1}{x}$

$\therefore \sin y = {x}^{-} 1$

Then, differentiating simplicity gives:

$\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = - {x}^{-} 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{{x}^{2} \cos y}$

And, using the trig odentity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have
$\cos y = \sqrt{1 - {\sin}^{2} y}$
$\therefore \cos y = \sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}$
$\therefore \cos y = \sqrt{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2}$
$\therefore \cos y = \sqrt{\frac{{x}^{2} - 1}{x} ^ 2}$
$\therefore \cos y = \frac{1}{x} \sqrt{{x}^{2} - 1}$

And, so substituting into the derivative above we get;
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{{x}^{2} \left(\frac{1}{x} \sqrt{{x}^{2} - 1}\right)}$
Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{x \sqrt{{x}^{2} - 1}}$